Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $p = \dfrac{-2z + 8}{9z + 45} \div \dfrac{4z + 32}{z^2 + 13z + 40} $
Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-2z + 8}{9z + 45} \times \dfrac{z^2 + 13z + 40}{4z + 32} $ First factor the quadratic. $p = \dfrac{-2z + 8}{9z + 45} \times \dfrac{(z + 5)(z + 8)}{4z + 32} $ Then factor out any other terms. $p = \dfrac{-2(z - 4)}{9(z + 5)} \times \dfrac{(z + 5)(z + 8)}{4(z + 8)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ -2(z - 4) \times (z + 5)(z + 8) } { 9(z + 5) \times 4(z + 8) } $ $p = \dfrac{ -2(z - 4)(z + 5)(z + 8)}{ 36(z + 5)(z + 8)} $ Notice that $(z + 8)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -2(z - 4)\cancel{(z + 5)}(z + 8)}{ 36\cancel{(z + 5)}(z + 8)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $p = \dfrac{ -2(z - 4)\cancel{(z + 5)}\cancel{(z + 8)}}{ 36\cancel{(z + 5)}\cancel{(z + 8)}} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $p = \dfrac{-2(z - 4)}{36} $ $p = \dfrac{-(z - 4)}{18} ; \space z \neq -5 ; \space z \neq -8 $